Longest Substring Without Repeating Characters

• Total Accepted: 167158
• Total Submissions: 735821
• Difficulty: Medium

Given a string, find the length of the longest substring without repeating characters.

Examples:

Given "abcabcbb", the answer is "abc", which the length is 3.

Given "bbbbb", the answer is "b"

Given "pwwkew", the answer is "wke", with the length of 3. Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

解題思路參考自： http://www.geeksforgeeks.org/length-of-the-longest-substring-without-repeating-characters/

 1 public class Num3 {
 2     /*
 3      * 方法一：暴力搜尋，複雜度 O（n^3）
 4      */
 5     public int lengthOfLongestSubstring(String s) {
 6         if(s == null || s.length() == 0){
 7             return 0 ;
 8         }
 9         String sub ;
10         for(int subLen = s.length() ; subLen > 0 ; subLen--){
11             for(int startIndex = 0 ; startIndex <= (s.length()-subLen) ; startIndex++){
12                 //列出所有子串，然後判斷子串是否滿足有重複
13                 if(startIndex != (s.length()-subLen)){
14                     sub = s.substring(startIndex, startIndex+subLen) ;
15                 }else{
16                     sub = s.substring(startIndex) ;
17                 }
18                 if(!isRepeat(sub)){
19                     return subLen ;
20                 }
21             }
22         }
23         
24         return 1 ;
25     }
26     
27     private boolean isRepeat(String s){
28         for(int i = 1 ; i < s.length(); i++){
29             if(s.substring(i).contains(s.substring(i-1, i))){
30                 return true ;
31             } 
32         }
33         return false ;
34     }
35     
36     /*
37      * 方法二：用hash的方法加上動態規劃求解
38      */
39     public int lengthOfLongestSubstring2(String s) {
40         if(s == null || s.length() == 0){
41             return 0 ;
42         }
43         int cur_len = 1 ;      //lenght of current substring
44         int max_len = 1 ;
45         int prev_index ;      //  previous index
46         int [] visited = new int [256] ;
47         char [] arr = s.toCharArray() ;
48         /* Initialize the visited array as -1, -1 is used to
49            indicate that character has not been visited yet. */
50         for(int i = 0 ; i < 256 ; i++){
51             visited[i] = -1 ;
52         }
53         /* Mark first character as visited by storing the index
54            of first   character in visited array. */
55         visited[arr[0]] = 0 ;
56         
57         /* Start from the second character. First character is
58            already processed (cur_len and max_len are initialized
59            as 1, and visited[arr[0]] is set */
60         for(int i = 1 ; i < arr.length ; i++){
61             prev_index = visited[arr[i]] ;
62             
63             /* If the current character is not present in the
64                already processed substring or it is not part of
65                the current NRCS, then do cur_len++ */
66             if(prev_index == -1 || i - cur_len > prev_index){
67                 cur_len++ ;
68             }else{
69                 /* Also, when we are changing the NRCS, we
70                    should also check whether length of the
71                    previous NRCS was greater than max_len or
72                    not.*/
73                 if(cur_len > max_len){
74                     max_len = cur_len ;
75                 }
76                 // update the index of current character
77                 cur_len = i - prev_index ;    
78             }
79             
80             visited[arr[i]] = i ;
81         }
82         
83         // Compare the length of last NRCS with max_len and
84         // update max_len if needed
85         if (cur_len > max_len){
86             max_len = cur_len ;
87         }
88         
89         return max_len ;
90         
91     }
92 
93 }